536 
THEOREMS IN RELATION TO CERTAIN SIGN-SYMBOLS. 
[546 
In particular a, b, c, d, e being independent, then a, b, c, bd, e (viz. any term d 
is replaced by its product by some other term b) is also independent; and by a similar 
transformation on the new series a, b, c, bd, e, and so on in succession we can pass 
from a given independent system a, b, c, d, e to any other independent system whatever. 
A similar but more general theorem is the following: let a, b, c, d, e be inde 
pendent, and l be equal to the product of all or any of these roots, but so that as 
regards, suppose (b, c, e), the number of these factors contained in l is even (or may 
be = 0), e.g. I is = a, or abc, &c., but it is not = ab, or bee, &c. 
Then a, lb, le, d, le is an independent system: to show this we must show that 
1 + al + b\ + cl + dl + e = l + al -\-lbl + lcl + dl + le, 
that is, 
1 a 1 d (1 lb 1 le 1 -\- le — 1 —{— & 1 —f- с 1 ~f- e) = 0, 
that is, 
l + al + d(l — lb + c+e + l 2 — 1 be + be + ce +1 3 — 1 bee) = 0, 
or, since l 2 =1, l 3 = l, this is 
(1 + ci) (1 + d) (l — 1 )(6 + c + 6 + bee) = 0, 
which is easily verified under the assumed conditions as to l, e.g. I = abc, 
(1 + a) l = abc + a?bc = abc + be = (1 + a) be, 
and the equation is 
and we in fact have 
that is, 
(1 +a)(l — l) = (l+a) (be — 1), 
(1 + a) (1 + d) (be — l)(ò + c + e + bee) = 0 ; 
be (b + c + e + bee) = c + 6 + ebc + e; 
(be — l)(b + c + e + bee) = 0. 
The proof is obviously quite general. 
All that precedes applies also to the columns. 
Now consider a square of 5 x 5 signs +; I say that, if this is independent as to 
its lines, it will be also independent as to its columns. 
To prove this consider any particular square, say 
a 
/? 
У 
8 
e 
a 
b 
+ 
- 
- 
+ 
- 
c 
d 
e